+296 Find the only real number that can be expressed in the form
\((a + bi)^3 - 107i\)
where \(i^2=-1\), and \(a\) and \(b\) are positive integers.
+129899 (a + bi)^3 =
a^3 + 3a^2 (bi) + 3a (bi)^2 + (bi)^3
a^3 + 3a^2bi - 3ab^2 - bi
We need to have
3a^2bi - bi = 107i
bi ( 3a^2 - 1) = 107i divide both sides by i
b ( 3a^2 - 1) = 107
Since 107 is prime.....then....
Let b = 1
So
3a^2 - 1 = 107
3a^2 = 108
a^2 = 36
a = 6
So
(6 + i)^3 - 107i =
6^3 + 3(6)^2 i + 3(6)(i)^2 - i - 107i =
216 + 108i - 18 - i - 107i =
(216 - 18) + (108i - 108i) =
198 = the real number
